Answer
$Pi_{He}$=0.201 atm
Work Step by Step
Data
$n_{He}$ = 0.0410
The volume of vessel = 5L = 5×$10^{−3}$m3
Temperature = 25∘C = 25 +273 = 298 K
$Pi_{He}$ =$ (n_{He}RT)/V$
=0.0410×8.314×2985×10−3
=20316.09 Nm2
$Pi_{He}$=0.201 atm
Chemistry (4th Edition)
by Burdge, Julia
Published by
McGraw-Hill Publishing Company
ISBN 10:
0078021529
ISBN 13:
978-0-07802-152-7
Chapter 10 - Section 10.5 - Gas Mixtures - Checkpoint - Page 451: 10.5.1
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